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Bingel [31]
3 years ago
13

Find the answer to 4.2 • 1.6 . Please helppp

Mathematics
1 answer:
borishaifa [10]3 years ago
7 0

4.2 x 1.6=6.72

Next time if you want, use the calculator

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3) Construct triangle PQR such that PQ = 6 cm, angle PQR =
raketka [301]

Step-by-step explanation:

kfucmc!fxlkdf?c

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8 0
3 years ago
Find the product.<br> (x2+6x +9)(x?) -<br> What’s the product
lesya692 [45]

Answer:

Step-by-step explanation:

(x² + 6x + 9)(x)= x²*x + 6x*x + 9*x = x³ + 6x²  + 9x

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3 years ago
Read 2 more answers
Solve the inequality 2n – 1 &lt; 39​
MatroZZZ [7]

Answer:

n < 20​

Step-by-step explanation:

2n – 1 < 39​

Add 1 to each side

2n – 1+1 < 39+1

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3 0
3 years ago
Expansion Numerically Impractical. Show that the computation of an nth-order determinant by expansion involves multiplications,
posledela

Answer:

  • number of multiplies is n!
  • n=10, 3.6 ms
  • n=15, 21.8 min
  • n=20, 77.09 yr
  • n=25, 4.9×10^8 yr

Step-by-step explanation:

Expansion of a 2×2 determinant requires 2 multiplications. Expansion of an n×n determinant multiplies each of the n elements of a row or column by its (n-1)×(n-1) cofactor determinant. Then the number of multiplies is ...

  mpy[n] = n·mp[n-1]

  mpy[2] = 2

So, ...

  mpy[n] = n! . . . n ≥ 2

__

If each multiplication takes 1 nanosecond, then a 10×10 matrix requires ...

  10! × 10^-9 s ≈ 0.0036288 s ≈ 0.004 s . . . for 10×10

Then the larger matrices take ...

  n=15, 15! × 10^-9 ≈ 1307.67 s ≈ 21.8 min

  n=20, 20! × 10^-9 ≈ 2.4329×10^9 s ≈ 77.09 years

  n=25, 25! × 10^-9 ≈ 1.55112×10^16 s ≈ 4.915×10^8 years

_____

For the shorter time periods (less than 100 years), we use 365.25 days per year.

For the longer time periods (more than 400 years), we use 365.2425 days per year.

8 0
3 years ago
Find the average montly loss of a conpany with a loss of 36,000 for one year. What's the equation?
algol13
36,00÷ 12= 3,000 so that would be the average monthly loss.
5 0
3 years ago
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